HDU 4681 String(DP)

题目链接

枚举A和B中每一段含有C的段，A的前面 后面和B前面后面，求最长公共子序。观察发现，可以预处理最长公共子序。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 int dp1[][],dp2[][];
 char s1[];
 char s2[];
 char s3[];
 char s4[];
 char s5[];
 int que1[][];
 int que2[][];
 int main()
 {
     int t,cas = ,i,j,n1,n2,k;
     int len1,len2,len3;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%s%s%s",s1,s2,s3);
         len1 = strlen(s1);
         len2 = strlen(s2);
         len3 = strlen(s3);
         for(i = ;i <= len1;i ++)
         {
             for(j = ;j <= len2;j ++)
             {
                 dp1[i][j] = dp2[i][j] = ;
             }
         }
         for(i = ;i <= len1;i ++)
         {
             for(j = ;j <= len2;j ++)
             {
                 if(s1[i-] == s2[j-])
                 dp1[i][j] = dp1[i-][j-] + ;
                 else
                 dp1[i][j] = max(dp1[i-][j],dp1[i][j-]);
             }
         }
         for(i = ;i < len1;i ++)
         {
             s4[i] = s1[len1-i-];
         }
         for(i = ;i < len2;i ++)
         {
             s5[i] = s2[len2-i-];
         }
         for(i = ;i <= len1;i ++)
         {
             for(j = ;j <= len2;j ++)
             {
                 if(s4[i-] == s5[j-])
                 dp2[i][j] = dp2[i-][j-] + ;
                 else
                 dp2[i][j] = max(dp2[i-][j],dp2[i][j-]);
             }
         }
         n1 = n2 = ;
         for(i = ;i < len1;i ++)
         {
             if(s1[i] == s3[])
             {
                 k = ;
                 for(j = i+;j < len1;j ++)
                 {
                     if(s1[j] == s3[k])
                     k ++;
                     if(k == len3) break;
                 }
                 if(j != len1)
                 {
                     que1[n1][] = i;
                     que1[n1][] = j;
                     n1 ++;
                 }
             }
         }
         for(i = ;i < len2;i ++)
         {
             if(s2[i] == s3[])
             {
                 k = ;
                 for(j = i+;j < len2;j ++)
                 {
                     if(s2[j] == s3[k])
                     k ++;
                     if(k == len3) break;
                 }
                 if(j != len2)
                 {
                     que2[n2][] = i;
                     que2[n2][] = j;
                     n2 ++;
                 }
             }
         }
         int ans = ;
         /*for(i = 1;i <= len1;i ++)
         {
             for(j = 1;j <= len2;j ++)
             {
                 printf("%d ",dp1[i][j]);
             }
             printf("\n");
         }*/
         /*for(i = 0;i < n1;i ++)
         {
             printf("%d %d\n",que1[i][0],que1[i][1]);
         }
         for(i = 0;i < n2;i ++)
         {
             printf("%d %d\n",que2[i][0],que2[i][1]);
         }*/
         for(i = ;i < n1;i ++)
         {
             for(j = ;j < n2;j ++)
             {
                 ans = max(ans,dp1[que1[i][]][que2[j][]] + dp2[len1-que1[i][]-][len2-que2[j][]-]);
             }
         }
         printf("Case #%d: %d\n",cas++,ans + len3);
     }
     return ;
 }